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WITCHER [35]
3 years ago
10

How do you express your terms in pi

Mathematics
2 answers:
Margarita [4]3 years ago
7 0
Good question I don’t know my own self
il63 [147K]3 years ago
4 0

Answer:

To express your answer in terms of pi, simply refrain from substituting pi's numerical value for its symbol in the equation. That way, your answer will look like xπ where x is whatever number you come up with, and π is simply a placeholder.

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C cubed minus the product of five and Q +4

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What is the variance?<br> What is the standard deviation, rounded to the nearest whole number?
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What is the variance? 84000

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3 years ago
Section 5.2 Problem 21:
Fittoniya [83]

Answer:

y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)] (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation am^2+bm+c=0 where the values of m are the roots:

y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i

Since the values of m are complex conjugate roots, then the general solution is y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)] where m=\alpha\pm\beta i.

Thus, the general solution for our given differential equation is y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)].

To account for both initial conditions, take the derivative of y(x), thus, y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]

Now, we can create our system of equations given our initial conditions:

y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3

y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2

We then solve the system of equations, which becomes easy since we already know that C_1=3:

-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}

Thus, our final solution is:

y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]

7 0
2 years ago
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